7. Zadatak

Najprije primijetimo da mora biti $x\ne0,\pm\frac12$.

$$\frac{2x+1}{3x(2x-1)}-\frac{2x-1}{7x(2x+1)}=\frac8{3(2x-1)(2x+1)}\;\Big/\;21x(2x-1)(2x+1)$$ $$(2x+1)\cdot7\cdot(2x+1)-3(2x-1)^2=8\cdot7x$$ $$7(4x^2+4x+1)-3(4x^2-4x+1)-56x=0$$ $$28x^2+28x+7-12x^2+12x-3-56x=0$$ $$16x^2-16x+4=0$$ $$4x^2-4x+1=0$$ $$(2x-1)^2=0.$$

No, na početku smo komentirali da mora biti $x\ne\frac12$ pa jednadžba nema rješenja.